Question

Let p0 be an equilateral triangle of area 10. Each side of p0 is trisected, and the corners are snipped off, creating a new polygon (in fact, a hexagon) p1. What is the area of p1? Now repeat the process to p1 i.E. Trisect each side and snip off the corners to obtain a new polygon p2. What is the area of p2? Now repeat this process infinitely often to create an object p. What is the area of p?

Answer 1

Answer:

Area of P = 3.57√3 (approximately)

Step-by-step explanation:

Given:

Area of equilateral triangle p1 = 10

Each side of p0 is trisected, and the corners are snipped off, creating a new polygon (in fact, a hexagon) p1.

Now repeat the process to p1 i.E. Trisect each side and snip off the corners to obtain a new polygon p2.

Now repeat this process infinitely often to create an object p.

To find:

Area of p when this process is repeated infinitely.

Solution:

Refer to the images attached below for the detailed solution. Now, we know that p0 is triangle whose sides are then trisected to give rise to p1 and this process is repeated infinitely.

Therefore, no. of sides in:

p0 = 3 sides (x3) (As each side is trisected)

p1 = 6 sides (x3)

p2 = 12 sides (x3) ..........and so on.

Therefore, to find out no. of sides, we use the following:

pn = 2p(n-1)

s(pn) = 2^n x 3

Therefore, p1 => hexagon = 6 x $\sqrt{3}$/4 x (10/3)^2 Or  $\sqrt{3}$/4(10)^2 - 3$\sqrt{3}$/4(10/3)^2

Now, p2 => 12 sided = 6 x 1/2 x (10/9)^2 x sin120 (Since, it is 6 x small triangles that are removed with 2 sides 10/9 each)

Now, we generalise,

Area (pn) = Area [p(n-1)] - sides (pn) x 1/2 (10/3^n)^2 x sin120

=> Area (pn) = Area [p(n-1)] - sides (pn) x $\sqrt{3}$/4 (10/3^n)^2

Now, denoting area by A and sides be S, we get:

A (pn) => A [p(n-1)] - s(pn) x $\sqrt{3}$/4 x (10/3^n)^2 = A [p(n-1)] - 75$\sqrt{3}$ x (2/9)^n

Therefore, when this is taken infinitely and added, we get:

A (pn) => A (p0) - c [2/9 + (2/9)^2 + ..... + (2/9)^n] = A (p0) - c * (sum of GP)

Now, if n tends to infinity, it becomes infinite GP,

A (pn) = A (p0) - c [(2/9) / (1 - 2/9)]

A(p∞) = A (p0) - 75√3 x 2/7

=> A(p∞) = (√3/4 x 10^2) - (75√3 x 2/7)

=> A(p∞) = √3 (25 - 21.42)

Therefore, A(p∞) = 3.57√3 (Approx.)

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